13—Vector Calculus 2 333I need only therˆcomponent of the curl because the surface integral uses only the normal (ˆr)
component. The surface integral of this has the area elementdA=r^2 sinθdθdφ.
∫
curlF~.dA~=
∫θ 00R^2 sinθdθ
∫ 2 π0dφ
1
Rsinθ
(
CRcos^2 φ2 sinθcosθ+BRθ^2 sinφ
)
=R^2
∫θ 00dθ
∫ 2 π0dφ 2 Ccos^2 φsinθcosθ
=R^22 Cπsin^2 θ 0 /2 =CR^2 πsin^2 θ 0
The other side of Stokes’ theorem is the line integral around the circle at angleθ 0.
∮
F~.d~`=
∫ 2 π0rsinθ 0 dφCrsinθcos^2 φ
=
∫ 2 π0dφCR^2 sin^2 θ 0 cos^2 φ
=CR^2 sin^2 θ 0 π (13.27)
and the two sides of the theorem agree. Check! Did I get the overall signs right? The direction of
integration around the loop matters. A further check: Ifθ 0 =π, the length of the loop is zero and
both integrals give zero as they should.
Conservative Fields
An immediate corollary of Stokes’ theorem is that if the curl of a vector field is zero throughout a region
then line integrals are independent of path in that region. To state it a bit more precisely, in a volume
for which any closed path can be shrunk to a point without leaving the region, if the curl of~vequals
zero, then
∫b
aF~.d~rdepends on the endpoints of the path, and not on how you get there.
To see why this follows, take two integrals from pointato pointb.
∫
1~v.d~r and
∫
2~v.d~r
a b
1
2
The difference of these two integrals is
∫1~v.d~r−
∫
2~v.d~r=
∮
~v.d~r
This equations happens because the minus sign is the same thing that you get by integrating in the
reverse direction. For a field with∇×~v= 0, Stokes’ theorem says that this closed path integral is
zero, and the statement is proved.
What was that fussy-sounding statement “for which any closed path can be shrunk to a point
without leaving the region?” Consider the vector field in three dimensions, written in rectangular and
cylindrical coordinates,